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M22-36

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M22-36  [#permalink]

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New post 16 Sep 2014, 01:17
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A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

87% (00:59) correct 13% (01:27) wrong based on 126 sessions

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Circles \(X\) and \(Y\) are concentric. If the radius of circle \(X\) is three times that of circle \(Y\), what is the probability that a point selected inside circle \(X\) at random will be outside circle \(Y\)?

A. \(\frac{1}{3}\)
B. \(\frac{\pi}{3}\)
C. \(\frac{\pi}{2}\)
D. \(\frac{5}{6}\)
E. \(\frac{8}{9}\)

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Re M22-36  [#permalink]

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New post 16 Sep 2014, 01:17
Official Solution:

Circles \(X\) and \(Y\) are concentric. If the radius of circle \(X\) is three times that of circle \(Y\), what is the probability that a point selected inside circle \(X\) at random will be outside circle \(Y\)?

A. \(\frac{1}{3}\)
B. \(\frac{\pi}{3}\)
C. \(\frac{\pi}{2}\)
D. \(\frac{5}{6}\)
E. \(\frac{8}{9}\)

We have to find the ratio of the area of the ring around the small circle to the area of the big circle. If \(y\) is the radius of the smaller circle, then the area of the bigger circle is \(\pi(3y)^2 = 9 \pi y^2\). The area of the ring \(= \pi(3y)^2 - \pi(y)^2 = 8 \pi y^2\). The ratio \(= \frac{8}{9}\).

Answer: E
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Re: M22-36  [#permalink]

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New post 21 Jan 2015, 06:49
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Hey,

Great that we saw how you do it using the actual variables.
However, I just used values.

For the radius of X = 6
For the radius of Y = 2

Then the area for X = 36π
and the area for Y = 4π

32/36 = 8/9.
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Re: M22-36  [#permalink]

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New post 26 Oct 2017, 15:20
I think the easiest way for me was:
\(\frac{πr^2}{π3r^2}\)

Use values Y=1, X=3Y=3

\(1^2 = 1, 3^2=9\), 1/9 chance it is inside the circle, or 8/9 chance it is outside.
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Re: M22-36  [#permalink]

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New post 25 Mar 2018, 20:02
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Bunuel What is concentric? Could you please give a pictorial solution? thanks
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Re: M22-36  [#permalink]

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New post 25 Mar 2018, 20:52
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Re M22-36  [#permalink]

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New post 03 Nov 2018, 06:16
I think this is a high-quality question and the explanation isn't clear enough, please elaborate.
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Re: M22-36  [#permalink]

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New post 13 Nov 2018, 21:58
Bunuel wrote:
Official Solution:

Circles \(X\) and \(Y\) are concentric. If the radius of circle \(X\) is three times that of circle \(Y\), what is the probability that a point selected inside circle \(X\) at random will be outside circle \(Y\)?

A. \(\frac{1}{3}\)
B. \(\frac{\pi}{3}\)
C. \(\frac{\pi}{2}\)
D. \(\frac{5}{6}\)
E. \(\frac{8}{9}\)

We have to find the ratio of the area of the ring around the small circle to the area of the big circle. If \(y\) is the radius of the smaller circle, then the area of the bigger circle is \(\pi(3y)^2 = 9 \pi y^2\). The area of the ring \(= \pi(3y)^2 - \pi(y)^2 = 8 \pi y^2\). The ratio \(= \frac{8}{9}\).

Answer: E



How did you deduce from the above question that we need to find the ratio of the areas?
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Re: M22-36   [#permalink] 13 Nov 2018, 21:58
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